Page:PoincareDynamiqueJuillet.djvu/45

On the other hand, we see that the following systems of quantities:

${\begin{array}{ccccccc}x,&&y,&&z,&&-r=t\\\\k_{0}X_{1},&&k_{0}Y_{1},&&k_{0}Z_{1},&&k_{0}T_{1}\\\\k_{0}\xi ,&&k_{0}\eta ,&&k_{0}\zeta ,&&k_{0}\\\\k_{1}\xi _{1},&&k_{1}\eta _{1},&&k_{1}\zeta _{1},&&k_{1}\end{array}}$

undergo the same linear substitutions when we apply the transformations of the Lorentz group. We are thus led to pose:

(9)

${\begin{cases}X_{1}=x{\frac {\alpha }{k_{0}}}+\xi \beta +\xi _{1}{\frac {k_{1}}{k_{0}}}\gamma ,\\\\Y_{1}=y{\frac {\alpha }{k_{0}}}+\eta \beta +\eta _{1}{\frac {k_{1}}{k_{0}}}\gamma ,\\\\Z_{1}=z{\frac {\alpha }{k_{0}}}+\zeta \beta +\zeta _{1}{\frac {k_{1}}{k_{0}}}\gamma ,\\\\T_{1}=-r{\frac {\alpha }{k_{0}}}+\beta +{\frac {k_{1}}{k_{0}}}\gamma ,\end{cases}}$

It is clear that if α, β, γ are invariants, X₁, Y₁, Z₁, T₁ satisfy the basic condition, that is to say, it will undergo, by the effect of the Lorentz transformations, a suitable linear substitution.

But for the equations (9) to be consistent, we must have:

$\sum X_{1}\xi -T_{1}=0,$

which, by replacing X₁, Y₁, Z₁, T₁ by their values (9) and multiplying by k₀², becomes:

(10)	$-A\alpha -\beta -C\gamma =0.\,$

What we want is, if we neglect the square of speed of light, the squares of the velocities ξ, etc., as well as the product of accelerations by the distances as we did above, so that the values of X₁, Y₁, Z₁ remain in conformity with the law of Newton.

We can take:

$\beta =0,\quad \gamma =-{\frac {A\alpha }{C}}.$

With the order of approximation adopted, we have:

$k_{0}=k_{1}=1,\quad C=1,\quad A=-r_{1}+\sum x\left(\xi _{1}-\xi \right),\quad B=-r_{1},$

$x=x_{1}+\xi _{1}t=x_{1}-\xi _{1}r\,$

The first equation (9) becomes:

$X_{1}=\alpha \left(x-A\xi _{1}\right)$

But if we neglect the square of ξ, we can replace Aξ₁ by -r₁ξ₁, or